I'm stuck on this question:

The original question is: find $\lim\limits_{x \to 0^+} x^{1/x}$.

This is equal to $\lim\limits_{x \to 0^+}e^{\frac{\ln(x)}{x}}$

I'm stuck at calculating $\lim\limits_{x \to 0^+}\frac{\ln(x)}{x}$.

I can't see how I can use L'Hopital's Rule or the Squeeze Theorem to solve this.

I know the final answer is $0$, so the limit in the exponent has to be negative infinity, but I don't understand why this is the answer.

  • $\begingroup$ Lhopital's rule does not apply to your situation. Note that $\ln x\to-\infty$ as $x\to 0^+$, and $x\to 0$ as $x\to 0^+$, thus $\ln x /x\to-\infty$ as $x\to 0^+$. $\endgroup$ – Frank Lu Mar 16 '15 at 23:28

There is no difficulty here :

$ \lim_{x \to 0} \; \ln(x) = -\infty $


$ \lim_{x \to 0} \; \dfrac{1}{x} = +\infty $

By product you have your $-\infty$. Used in exp you find the final result.

  • $\begingroup$ sure its infinity * infinity.... silly me... $\endgroup$ – AK_ Mar 16 '15 at 23:37

L'Hospital does not apply: as $x\to0^+$, we have $\log(x)\to-\infty$ and $x\to0^+$.

Ignoring the signs at first, what happens when the numerator gets big and the denominator gets small?

Next, apply the signs.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.